What is the integral from 0 to pi/4 of t^2sin(2t) ?

The integral from 0 to pi/4 of t^2sin(2t) is pi/8-1/4

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integral from 0 to pi/4 of t^2sin(2t)

\int_{0}^{\frac{π}{4}} t^{2} sin (2 t) d t

\frac{π}{8} - \frac{1}{4}

Decimal

0.14269 \dots

Solution steps

\int_{0}^{\frac{π}{4}} t^{2} sin (2 t) d t

Apply Integration By Parts

: [\frac{1}{2} (- t^{2} cos (2 t) - 2 \cdot \int - t cos (2 t) d t)]_{0}^{\frac{π}{4}}

= [\frac{1}{2} (- t^{2} cos (2 t) - 2 \cdot \int - t cos (2 t) d t)]_{0}^{\frac{π}{4}}

\int - t cos (2 t) d t = - \frac{1}{4} (2 t sin (2 t) + cos (2 t))

= [\frac{1}{2} (- t^{2} cos (2 t) - 2 (- \frac{1}{4} (2 t sin (2 t) + cos (2 t))))]_{0}^{\frac{π}{4}}

Simplify

[\frac{1}{2} (- t^{2} cos (2 t) - 2 (- \frac{1}{4} (2 t sin (2 t) + cos (2 t))))]_{0}^{\frac{π}{4}} : [\frac{1}{4} (- 2 t^{2} cos (2 t) + 2 t sin (2 t) + cos (2 t))]_{0}^{\frac{π}{4}}

= [\frac{1}{4} (- 2 t^{2} cos (2 t) + 2 t sin (2 t) + cos (2 t))]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{π}{8} - \frac{1}{4}

= \frac{π}{8} - \frac{1}{4}

\int_{- 2}^{2} 2 x - 4 d x

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\int_{0}^{2} x (8 x - 2 x) d x

\int_{0}^{2} (6 - 3 x) d x

\int_{0}^{3} \frac{x ^{2}}{x ^{3} + 2} d x

What is the integral from 0 to pi/4 of t^2sin(2t) ?
The integral from 0 to pi/4 of t^2sin(2t) is pi/8-1/4